If Z is an idempotent matrix, then `(I+Z)^(n)`

Options

A. `I+2^(n)Z`
B. `I+(2^(n)-1) Z`
C. `I-(2^(n)-1)Z`
D. none of these

Explanation

To solve the problem, we need to find the expression for \((I + Z)^n\) where \(Z\) is an idempotent matrix. An idempotent matrix \(Z\) satisfies the property \(Z^2 = Z\).

### Step-by-Step Solution:

1. **Understanding Idempotent Matrix**:
Since \(Z\) is idempotent, we have:
\[ Z^2 = Z, \quad Z^3 = Z, \quad \ldots, \quad Z^n = Z \text{ for any } n \geq 1. \]

2. **Applying the Binomial Theorem**:
We can expand \((I + Z)^n\) using the binomial theorem:
\[ (I + Z)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} Z^k. \]

3. **Simplifying the Terms**:
Since \(I^{n-k} = I\) for any \(k\), we can simplify the expression:
\[ (I + Z)^n = \sum_{k=0}^{n} \binom{n}{k} I Z^k = I \sum_{k=0}^{n} \binom{n}{k} Z^k. \]

4. **Replacing \(Z^k\)**:
For \(k \geq 1\), since \(Z\) is idempotent, we have:
\[ Z^k = Z \text{ for } k \geq 1. \]
Thus, the sum can be rewritten as:
\[ (I + Z)^n = I \cdot \binom{n}{0} + I \cdot \left( \sum_{k=1}^{n} \binom{n}{k} Z \right). \]

5. **Calculating the Coefficients**:
The first term is:
\[ \binom{n}{0} = 1. \]
The second term becomes:
\[ Z \sum_{k=1}^{n} \binom{n}{k} = Z \left( 2^n - 1 \right), \]
since \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\).

6. **Final Expression**:
Therefore, we can combine these results:
\[ (I + Z)^n = I + (2^n - 1)Z. \]

### Conclusion:
Thus, the final expression for \((I + Z)^n\) is:
\[ (I + Z)^n = I + (2^n - 1)Z. \]